BWhat Do You Get When You Multiply An Object's mass times the acceleration?

Hello guys I have a quick question and please help me! Thanks!

xenn [34]

Answer: 4 herz is the answer!

Explanation:

6 0

3 years ago

Read 2 more answers

A long string is pulled so that the tension in it increases by a factor of four. If the change in length is negligible, by what

rusak2 [61]

To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is

$v = \sqrt{\frac{T}{\mu}}$

Here

v = Wave speed

T = Tension

$\mu$ = Linear mass density

From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension

$v \propto \sqrt{T}$

Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion

$v \propto \sqrt{4} = 2$

The factor that the wave speed change is 2.

3 0

3 years ago

A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m

olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

$v = v_0 -gt$

Here,

v = Final velocity

$v_0$ = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

$-30 = v_0 -9.8(4)$

$v_0 = 9.2m/s$

Now the position can be calculated as,

$y = h +v_0t -\frac{1}{2}gt^2$

When it has the ground, y=0 and the time is t=4s,

$0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2$

$h = 41.6m$

Therefore the cliff was initially to 41.6m from the ground

7 0

3 years ago

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago

A 2.4-m high 200-m2 house is maintained at 22Â°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch

kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is <u>$0.50</u> for 10 hours.

Option: a

<u>Explanation</u>:

"Energy discharged by air in every hour" can be determined by,

$\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

$\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}$

$\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Density of air } \rho=1.20 \times 200 \times 2.4$

$\text { Density of air } \rho=576 \mathrm{kg}$

∆T = 10 hours

$\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}$